Relation between Moore-penrose pseudoinverse and $A^T$?

Question

I have the following question:

If $Ax=b$ where $A \in R^{m \times n}$, then we can get $x = A'b$ where $A'$ is the moore-pensrose psuedoinverse. Is it true that $x$ is unique if $x^Ty = 0 \forall y \in N(A)$ and $$x \in R(A^T) = N^{\perp}(A)?$$

Thanks in advance for the help.

Answer 1

The relation between the spaces is given by:

What you can say is if $ \boldsymbol{b} \in \mathcal{R} \left( \boldsymbol{A} \right) $ then there is a solution to the equation.

All solution will be in the form $ \boldsymbol{x}_{r} + \boldsymbol{x}_{n} $ where $ \boldsymbol{x}_{r} \in \mathcal{R} \left( \boldsymbol{A}^{T} \right) $ and $ \boldsymbol{x}_{n} \in \mathcal{N} \left( \boldsymbol{A} \right) $.

The only way for this set of solution to be a singleton is in case the null space is empty.

What can be shown is that the Pseudo Inverse using the SVD gives the least norm solution. Namely it gives back $ \boldsymbol{x}_{r} $.

Here is a sketch of a proof for that:

Answer 2

I haven't tried to parse your formula, but here is another picture showing what the Moore-Penrose pseudoinverse does.

Here's a verbal description: The input to the pseudoinverse is a vector $y$ in the codomain. Project it orthogonally to the image, i.e., to the point $\bar{y}$ closest to $y$ in the image. Pick any vector $\bar{x}$ in the preimage of $\bar{y}$ and project it onto the orthogonal complement of the kernel. The resulting vector (labeled $L^+(y)$ in the picture, which does not depend on which vector you chose in the preimage, is the output of the pseudoinverse.